std2-std1-common

Financial Maths, STD1 F3 2025 HSC 27

The graph shows the salvage value of a car over 5 years.

The salvage values are based on the declining-balance method.

By what amount will the car’s value depreciate during the 10th year? (4 marks)

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$$1476.40$

Show Worked Solution

$\text{Find}\ r:$

$\text{When}\ \ n=1, \ S=$44\ 000\ \ \text{(see graph)}$

$S$	$=V_0(1-r)^n$
$44\ 000$	$=55\ 000(1-r)^1$
$\dfrac{44\ 000}{55\ 000}$	$=1-r$
$1-r$	$=0.8$
$r$	$=1-0.8=0.20$

$\text{Find \(S$ when}\ \ n=9\ \ \text{and}\ \ n=10:\)

$S_9=55\ 000(1-0.20)^{9}=$7381.97504$

$S_{10}=55\ 000(1-0.20)^{10}=$5905.5800$

$S_9-S_{10}=$7381.9750-$5905.580=$1476.40\ \text{(nearest cent)}$

$\therefore\ \text{The car’s value will depreciate by \$1476.40 in the 10th year.}$

Financial Maths, STD1 F3 2025 HSC 24

A used car has a sale price of $24 200. In addition to the sale price, the following costs are charged:

transfer of registration $50
stamp duty which is calculated at $3 for every $100, or part thereof, of the sale price.

Kat borrows the total amount to be paid for the car, including transfer of registration and stamp duty. Simple interest at the rate of 6.8% per annum is charged on the loan. The loan is to be repaid in equal monthly repayments over 3 years.

Calculate Kat’s monthly repayment. (5 marks)

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$$835.31$

Show Worked Solution

$\text{Stamp Duty} =\dfrac{24\ 200}{100}\times 3=$726$

$\text{Total Cost}\ $	$\text{= Price + Transfer + Stamp Duty}$
	$\text{= }24\ 200+50+726$
	$\text{= }$24\ 976$

$\text{Interest}=Prn=24\,976\times 0.068\times 3=$5095.104$

$\text{Loan amount}\ $	$\text{= Total Cost + Interest}$
	$\text{= }24\ 976+5095.104$
	$\text{= }$30\ 071.104$

$\text{3 years}= 3 \times 12=36\ \text{months}$

$\text{Monthly repayment}$	$=\dfrac{30\ 071.104}{36}$
	$=$835.308444$
	$\approx $835.31$

Algebra, STD1 A3 2025 HSC 23

It costs $465 to register a passenger car and $350 to register a motorcycle.

Let	$P$	$\ =\ \text{the number of passenger cars, and}$
	$B$	$\ =\ \text{the number of motorcycles}$

Write TWO linear equations that represent the relationship below.

There are 11 times as many passenger cars as motorcycles.
The total registration fees for passenger cars and motorcycles is $494 million. (2 marks)

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$\text{Equation 1: }\ P=11B$

$\text{Equation 2: }\ 465P+350B=494\ 000\ 000$

Show Worked Solution

$\text{Equation 1: }\ P=11B$

$\text{Equation 2: }\ 465P+350B=494\ 000\ 000$

Financial Maths, STD1 F3 2025 HSC 21

A credit card has an interest-free period of 45 days from and including the date of purchase. Interest is charged on purchases made, compounding daily at a rate of 13.74% per annum, from and including the day following the interest-free period.

Concert tickets were purchased for a total of $392 using this credit card.

Full payment was made on the 68th day from the date of purchase. There were no other purchases on this credit card.

What was the total interest charged when the account was paid in full? (3 marks)

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$\text{Interest charged}\ =\$ 3.41$

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$\text{Day 1-45: no interest is charged}$

$\text{Day 46-68: interest charged (23 days)}$

$\text{Daily interest rate}=\dfrac{13.74}{365} \%=\dfrac{13.74}{365 \times 100}$

$\text{Amount owing}=392\left(1+\dfrac{13.74}{365 \times 100}\right)^{23}=\$ 395.41$

$\text{Interest charged}=395.41-392=\$ 3.41$

Measurement, STD1 M4 2025 HSC 17

Paint is sold in two sizes at a local shop.

Four-litre cans at $90 per can
Ten-litre cans at $205 per can

Mina needs to purchase 80 litres of paint.

Calculate the amount of money Mina will save by purchasing only ten-litre cans of paint rather than only four-litre cans of paint. (2 marks)

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$\text{Saving}=$160$

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$\text{4 litre cans needed}\ =\dfrac{80}{4}=20$

$\text{Cost using 4 litre cans}\ =20\times $90=$1800$

$\text{10 litre cans needed}\ =\dfrac{80}{10}=8$

$\text{Cost using 10 litre cans}\ =8\times $205=$1640$

$\therefore\ \text{Saving}\ =$1800-$1640=$160$

Measurement, STD1 M5 2025 HSC 10 MC

The ratio of the dimensions of a model car to the dimensions of an actual car is $1:64$. The actual car has a length of 4.9 m.

What is the length of the model car in cm, correct to 1 decimal place?

3.1
7.7
13.1
59.1

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$B$

Show Worked Solution

$\text{Actual length}=4.9\ \text{m}\ =490\ \text{cm}$

$\therefore\ \text{Model car length}\ =490\times\dfrac{1}{64}=7.65625\approx 7.7\ \text{cm}$

$\Rightarrow B$

Probability, STD1 S2 2025 HSC 8 MC

A spinner made up of 4 colours is spun 100 times. The frequency histogram shows the results.

Which of these spinners is most likely to give the results shown?

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$A$

Show Worked Solution

$P(\text{White})$	$=\dfrac{50}{100}=\dfrac{1}{2}$
$P(\text{Red})$	$=\dfrac{25}{100}=\dfrac{1}{4}$
$P(\text{Yellow})$	$=\dfrac{15}{100}=\dfrac{3}{20}$
$P(\text{Green})$	$=\dfrac{10}{100}=\dfrac{2}{20}=\dfrac{1}{10}$

$\text{Eliminate Options B and D as white}\ \neq \dfrac{1}{2}\ \text{of spinner.}$

$\text{Eliminate Option C as red}\ \neq \dfrac{1}{4}\ \text{of spinner.}$

$\Rightarrow A$

Networks, STD1 N1 2025 HSC 6 MC

The network shows the distances, in kilometres, along a series of roads that connect towns.

What is the value of the largest weighted edge included in the minimum spanning tree for this network?

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$C$

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$\text{Minimum spanning tree:}$

$\text{Using Kruskal’s algorithm:}$

$\text{Edge 1 = 4, Edge 2/3 = 4, Edge 4 = 6, Edge 5 = 9}$

$\therefore\ \text{The largest weighted edge in the MST = 9.}$

$\Rightarrow C$

Financial Maths, STD1 F3 2023 HSC 30

A plumber leases equipment which is valued at $60 000.

The salvage value of the equipment at any time can be calculated using either of the two methods of depreciation shown in the table.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Method of depreciation} \rule[-1ex]{0pt}{0pt} & \textit{Rate of depreciation} \\
\hline
\rule{0pt}{2.5ex} \text{Straight-line method} \rule[-1ex]{0pt}{0pt} & \text{\$3500 per annum} \\
\hline
\rule{0pt}{2.5ex} \text{Declining-balance method} \rule[-1ex]{0pt}{0pt} & \text{12% per annum} \\
\hline
\end{array}

Under which method of depreciation would the salvage value of the equipment be lower at the end of 3 years? Justify your answer with appropriate mathematical calculations. (3 marks)

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$\text{Straight-line method:}$

$S$	$=V_0-Dn$
	$=60\ 000-3500\times 3$
	$=$49\ 500$

$\text{Declining-balance method:}$

$S$	$=V_0(1-r)^n$
	$=60\ 000(1-0.12)^3$
	$=60\ 000(0.88)^3$
	$=$40\ 888.32$

$\text{Salvage value is lower for the declining-balance method.}$

Show Worked Solution

$\text{Straight-line method:}$

$S$	$=V_0-Dn$
	$=60\ 000-3500\times 3$
	$=$49\ 500$

$\text{Declining-balance method:}$

$S$	$=V_0(1-r)^n$
	$=60\ 000(1-0.12)^3$
	$=60\ 000(0.88)^3$
	$=$40\ 888.32$

$\text{Salvage value is lower for the}$

$\text{declining-balance method.}$

♦ Mean mark 48%.

Measurement, STD1 M3 2023 HSC 29

The diagram shows the location of three places $X$, $Y$ and $C$.

$Y$ is on a bearing of 120° and 15 km from $X$.

$C$ is 40 km from $X$ and lies due west of $Y$.

$P$ lies on the line joining $C$ and $Y$ and is due south of $X$.

Find the distance from $X$ to $P$. (2 marks)

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What is the bearing of $C$ from $X$, to the nearest degree? (2 marks)

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$7.5\ \text{km}$
$259^{\circ}$

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a. $\text{In}\ \Delta XPY:$

$\angle PXY=180-120=60^{\circ}$

$\cos 60^{\circ}$	$=\dfrac{XP}{15}$
$XP$	$=15\times \cos 60^{\circ}$
	$=7.5\ \text{km}$

♦♦ Mean mark (a) 24%.

b. $\text{In}\ \Delta XPC:$

$\text{Let}\ \theta = \angle CXP$

$\cos \theta$	$=\dfrac{7.5}{40}$
$\theta$	$=\cos^{-1} \Big(\dfrac{7.5}{40}\Big)$
	$=79.193…$
	$=79^{\circ}\ \text{(nearest degree)}$

$\text{Bearing}\ C\ \text{from}\ X$	$=180+79$
	$=259^{\circ}$

♦♦♦♦ Mean mark (b) 9%.

Algebra, STD1 A3 2023 HSC 26

Electricity provider $A$ charges 25 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $40.

Complete the table showing Provider $A$ 's monthly charges for different levels of electricity usage. (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ 0\ \ \ \ \ & \ \ \ 400\ \ \ & \ \ 1000\ \ \\
\hline
\rule{0pt}{2.5ex} \text{Monthly charge (\$)} \rule[-1ex]{0pt}{0pt} & 40 & & 290 \\
\hline
\end{array}

Provider $B$ charges 35 cents per kWh, with no fixed monthly charge. The graph shows how Provider $B$ 's charges vary with the amount of electricity used in a month.

On the grid on above, graph Provider A's charges from the table in part (a). (1 mark)
Use the two graphs to determine the number of kilowatt hours per month for which Provider $A$ and Provider $B$ charge the same amount. (1 mark)

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A customer uses an average of 800 kWh per month.
Which provider, $A$ or $B$, would be the cheaper option and by how much? (2 marks)

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c. $400\text{ kWh}$

d. $A\text{ is cheaper by }$40.$

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♦♦ Mean mark (b) 40%.

c. $\text{Same charge when Provider }A = \text{Provider } B \text{ i.e. where the lines intersect}$

$=400\text{ kWh (see graph above)}$

d. $\text{When kWh}= 800, \ \ A=$240 \text{ and } B=$280$

$\therefore A\text{ is cheaper by }$40.$

Networks, STD1 N1 2023 HSC 15

The table shows some of the flight distances (rounded to the nearest 10 km between various Australian cities.

Use the information in the table to complete the network diagram where the edges are labelled with distances. (2 marks)

Mahsa wants to travel from Hobart to Darwin. She wants to change planes only once.
Using the network diagram, calculate how many kilometres she will travel by plane. (1 mark)

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b. $4190\text{ km}$

Show Worked Solution

b. $\text{Changing planes only once}\Rightarrow H\rightarrow S\rightarrow D$

$\text{Kilometres travelled}=1040+3150=4190\text{ km}$

Measurement, STD1 M5 2023 HSC 9 MC

A bag contains 150 jelly beans. Some of them are red and the rest are blue. The ratio of red to blue jelly beans is 2 : 3.

Sophie eats 10 of each colour.

What is the new ratio of red to blue jelly beans?

$2 : 3$
$4 : 9$
$5 : 8$
$11 : 17$

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$C$

Show Worked Solution

$\text{Original ratio}\ = 2:3$

$\text{5 parts = 150}\ \Rightarrow\ \text{1 part}\ = 150/5=30$

$\text{Original ratio (by number)}\ = 2 \times 30:3 \times 30 = 60:90$

$\text{After eating 10 of each colour:}$

$\text{Ratio}\ = 50:80 = 5:8$

$\Rightarrow C$

♦♦ Mean mark 34%.

Financial Maths, STD1 F1 2023 HSC 7 MC

An item was purchased for a price of $$880$, including $10\%$ GST.

What is the amount of GST included in the price?

$$8.00$
$$8.80$
$$80.00$
$$88.00$

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$C$

Show Worked Solution

$\text{Let}\ C =\text{Original cost}$

$C+0.1 \times C$	$=880$
$1.1C$	$=880$
$C$	$=\dfrac{880}{1.1}$
	$=$800$

$\therefore \text{GST}=800\times 0.1=$80$

$\Rightarrow C$

♦ Mean mark 15%.

Financial Maths, STD1 F1 2023 HSC 6 MC

A delivery truck was valued at $$65 000$ when new. The value of the truck depreciates at a rate of $22$ cents per kilometre travelled.

What is the value of the truck after it has travelled a total distance of $132 600$ km?

$$35 828$
$$29 172$
$$14 872$
$$14 300$

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$A$

Show Worked Solution

$\text{Value}$	$=65\ 000-132\ 600\times\frac{22}{100}$
	$=$35\ 828$

$\Rightarrow A$

Financial Maths, STD1 F1 2023 HSC 2 MC

An amount of $2500 is invested at a simple interest rate of 3% per annum.

How much interest is earned in the first two years?

$75
$150
$2575
$2652

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$B$

Show Worked Solution

$I$	$=Prn$
	$=2500 \times\ \dfrac{3}{100} \times\ 2$
	$=$150$

$\Rightarrow B$

\(S\)	\(=V_0(1-r)^n\)
\(44\ 000\)	\(=55\ 000(1-r)^1\)
\(\dfrac{44\ 000}{55\ 000}\)	\(=1-r\)
\(1-r\)	\(=0.8\)
\(r\)	\(=1-0.8=0.20\)

\(\text{Total Cost}\ \)	\(\text{= Price + Transfer + Stamp Duty}\)
	\(\text{= }24\ 200+50+726\)
	\(\text{= }$24\ 976\)

\(\text{Loan amount}\ \)	\(\text{= Total Cost + Interest}\)
	\(\text{= }24\ 976+5095.104\)
	\(\text{= }$30\ 071.104\)

\(\text{Monthly repayment}\)	\(=\dfrac{30\ 071.104}{36}\)
	\(=$835.308444\)
	\(\approx $835.31\)

Let	\(P\)	\(\ =\ \text{the number of passenger cars, and}\)
	\(B\)	\(\ =\ \text{the number of motorcycles}\)

\(P(\text{White})\)	\(=\dfrac{50}{100}=\dfrac{1}{2}\)
\(P(\text{Red})\)	\(=\dfrac{25}{100}=\dfrac{1}{4}\)
\(P(\text{Yellow})\)	\(=\dfrac{15}{100}=\dfrac{3}{20}\)
\(P(\text{Green})\)	\(=\dfrac{10}{100}=\dfrac{2}{20}=\dfrac{1}{10}\)

\(S\)	\(=V_0-Dn\)
	\(=60\ 000-3500\times 3\)
	\(=$49\ 500\)

\(S\)	\(=V_0(1-r)^n\)
	\(=60\ 000(1-0.12)^3\)
	\(=60\ 000(0.88)^3\)
	\(=$40\ 888.32\)

\(S\)	\(=V_0-Dn\)
	\(=60\ 000-3500\times 3\)
	\(=$49\ 500\)